A thin converging lens of focal length $f=25 \mathrm{~cm}$ forms the image of an object on a screen placed at a distance of $75 \mathrm{~cm}$ from the lens. The screen is moved closer to the lens by a distance of $25 \mathrm{~cm}$. The distance through which the object has to be shifted, so that its image on the screen in sharp again is

• A

  $37.5 \mathrm{~cm}$
• B

  $16.25 \mathrm{~cm}$
• Correct answer
  C

  $12.5 \mathrm{~cm}$
• D

  $13.5 \mathrm{~cm}$

Solution

According to the first condition,
$$ \begin{aligned} f & =25 \mathrm{~cm}, v=75 \mathrm{~cm} \\ u & =? \\ \frac{1}{f} & =\frac{1}{v}-\frac{1}{u} \\ \frac{1}{25} & =\frac{1}{75}-\frac{1}{u} \\ \frac{1}{u} & =\frac{1}{75}-\frac{1}{25} \\ \frac{1}{u} & =\frac{1-3}{75} \\ u & =-\frac{75}{2}=-37.5 \mathrm{~cm} \end{aligned} $$ According to the second condition
$$ \begin{aligned} v_1=50 \mathrm{~cm}, f & =25 \mathrm{~cm}, u_1=? \\ \frac{1}{f} & =\frac{1}{v_1}-\frac{1}{u_1} \\ \frac{1}{25} & =\frac{1}{50}-\frac{1}{u_1} \\ \frac{1}{u_1} & =\frac{1}{50}-\frac{1}{25} \\ \Rightarrow \quad \frac{1}{u_1} & =\frac{1-2}{50} \\ \Rightarrow \quad u_1 & =-50 \mathrm{~cm} \end{aligned} $$ So, the screen is sharp again is $$ \begin{aligned} \Delta u & =u_1-u \\ & =50-37.5 \\ & =12.5 \mathrm{~cm} \end{aligned} $$

Hence, the correct answer is option: 'C'.